If you had gone with your first idea, [itex]t= x^2[/itex] so that dt= 2x dx and (1/2)dt= x dx, you would have got [itex](1/2)\int sin(t)/cos^3(t) dt[/itex]. Now, it should be clear, with that "sin(t)dt" in the numerator, you can let y= cos(t), so that dy= sin(t) dt, and the integral becomes [itex](1/2)\int dy/y^3= (1/2)\int y^{-3}dy[/itex].

Essentially, you are doing the same thing in two steps rather than one.

If you had gone with your first idea, [itex]t= x^2[/itex] so that dt= 2x dx and (1/2)dt= x dx, you would have got [itex](1/2)\int sin(t)/cos^3(t) dt[/itex]. Now, it should be clear, with that "sin(t)dt" in the numerator, you can let y= cos(t), so that dy= sin(t) dt, and the integral becomes [itex](1/2)\int dy/y^3= (1/2)\int y^{-3}dy[/itex].

Essentially, you are doing the same thing in two steps rather than one.

Sir correct me if i am wrong but i think there is a mistake in the method you have done which i first stated.

if we took [itex]t= x^2[/itex] and dt =2x dx we cannot substitute that in the equation because ##x^2## is within the sine function and dx is outside???